Question: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{q^3 - 14q^2 + 49q}{3q^2 - 30q + 63}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {q(q^2 - 14q + 49)} {3(q^2 - 10q + 21)} $ $ k = \dfrac{q}{3} \cdot \dfrac{q^2 - 14q + 49}{q^2 - 10q + 21} $ Next factor the numerator and denominator. $ k = \dfrac{q}{3} \cdot \dfrac{(q - 7)(q - 7)}{(q - 7)(q - 3)}$ Assuming $q \neq 7$ , we can cancel the $q - 7$ $ k = \dfrac{q}{3} \cdot \dfrac{q - 7}{q - 3}$ Therefore: $ k = \dfrac{ q(q - 7)}{ 3(q - 3)}$, $q \neq 7$